I Showing ##k[x_1,\ldots,x_n]/\mathfrak{a}## is finite dimensional

[fresh_42]

@fresh_42 oh right, right. I was too quick and hopeful about the sum of ideals. Anyways, I will post the questions about the other direction here tomorrow. I think it is late for you. I am in eastern standard time -5:00 hrs. i think you are more east than me.

Six hours to Detroit.

 

[elias001]

@fresh_42 For the other direction, it says:
Assume that for each $i$, there is some $i$ for which $x^d_i\in\mathfrak{a}$.

Then, for all $m_i\geq d_i$,

we have $x_i^{m_i}=x^{m_i-d_i} x_i^{d_i}\in\mathfrak{a}$.

In particular, take a monomial $x_1^{m_1}\cdots x_n^{m_n}$.

If there exists at least one $i$ such that $m_i\geq d_i$,

then $x_1^{m_1}\cdots x_n^{m_n}\in\mathfrak{a}$.

Now, an element of the quotient has the form $P=\sum_{m_1,\ldots,m_n\geq 0}a_{m_1,\ldots,m_n} x_1^{m_1}\cdots x_n^{m_n}+\mathfrak{a}.$

As you notice, for all $Q\in \mathfrak{a}$,

we have $P+Q+\mathfrak{a}=P+\mathfrak{a}.$

The previous point then shows that $P=\sum_{m_1\leq d_1-1,\ldots,m_n\leq d_n-1}a_{m_1,\ldots,m_n} x_1^{m_1}\cdots x_n^{m_n}+\mathfrak{a}.$

Hence, the classes $x_1^{m_1}\cdots x_n^{m_n}, m_i\leq d_i-1$

for all $i$, span the vector space $k[x_1,\ldots,x_n]/\mathfrak{a}$.

Since this family is finite, it follows that this vector space is finite dimensional (of dimension $\leq d_1\cdots d_n$).

Note that we didn't use the fact that $\mathfrak{a}$ is generated by monomials here.


I don't understand the sentence: "The previous point then shows that $P=\sum_{m_1\leq d_1-1,\ldots,m_n\leq d_n-1}a_{m_1,\ldots,m_n} x_1^{m_1}\cdots x_n^{m_n}+\mathfrak{a}.$

Hence, the classes $x_1^{m_1}\cdots x_n^{m_n}, m_i\leq d_i-1$

for all $i$, span the vector space $k[x_1,\ldots,x_n]/\mathfrak{a}$."

In particular, I am not seeing how the inequalities: $m_1\leq d_1-1,\ldots,m_n\leq d_n-1$

is arrived at since the author let $x^{m_i}=x^{m_i-d_i}x^{d_i}$. So $m_i=m_i-d_i+d_i\geq d_i$.

Also, how does the

"the classes $x_1^{m_1}\cdots x_n^{m_n}, m_i\leq d_i-1$"

span the vector space $k[x_1,\ldots,x_n]/\mathfrak{a}$?

Finally, when the author chooses a particular monomial then says "there exists at least one $i$ such that $m_i\geq d_i$" make the proof to feel like he is going for a proof by contradiction. Since in the beginning of the proof, it assumes "for every $i$, now, there is at least one $i$.

 

[fresh_42]

@fresh_42 For the other direction, it says:
Assume that for each $i$, there is some $i$ for which $x^d_i\in\mathfrak{a}$.

Then, for all $m_i\geq d_i$,

we have $x_i^{m_i}=x^{m_i-d_i} x_i^{d_i}\in\mathfrak{a}$.

In particular, take a monomial $x_1^{m_1}\cdots x_n^{m_n}$.

If there exists at least one $i$ such that $m_i\geq d_i$,

then $x_1^{m_1}\cdots x_n^{m_n}\in\mathfrak{a}$.

Now, an element of the quotient has the form $P=\sum_{m_1,\ldots,m_n\geq 0}a_{m_1,\ldots,m_n} x_1^{m_1}\cdots x_n^{m_n}+\mathfrak{a}.$

As you notice, for all $Q\in \mathfrak{a}$,

we have $P+Q+\mathfrak{a}=P+\mathfrak{a}.$

The previous point then shows that $P=\sum_{m_1\leq d_1-1,\ldots,m_n\leq d_n-1}a_{m_1,\ldots,m_n} x_1^{m_1}\cdots x_n^{m_n}+\mathfrak{a}.$

Hence, the classes $x_1^{m_1}\cdots x_n^{m_n}, m_i\leq d_i-1$

for all $i$, span the vector space $k[x_1,\ldots,x_n]/\mathfrak{a}$.

Since this family is finite, it follows that this vector space is finite dimensional (of dimension $\leq d_1\cdots d_n$).

Note that we didn't use the fact that $\mathfrak{a}$ is generated by monomials here.


I don't understand the sentence: "The previous point then shows that $P=\sum_{m_1\leq d_1-1,\ldots,m_n\leq d_n-1}a_{m_1,\ldots,m_n} x_1^{m_1}\cdots x_n^{m_n}+\mathfrak{a}.$

Hence, the classes $x_1^{m_1}\cdots x_n^{m_n}, m_i\leq d_i-1$

for all $i$, span the vector space $k[x_1,\ldots,x_n]/\mathfrak{a}$."

In particular, I am not seeing how the inequalities: $m_1\leq d_1-1,\ldots,m_n\leq d_n-1$

is arrived at since the author let $x^{m_i}=x^{m_i-d_i}x^{d_i}$. So $m_i=m_i-d_i+d_i\geq d_i$.

These were two different considerations.

Try the trick with $z$. Imagine that you replaced every occurrence of $x_i^{d_i}$ by $z$. Then you get a polynomial in $z$ with coefficients of the form $x_1^{m_1}\cdots x_n^{m_n}, m_i\leq d_i-1.$ Since $z\in \mathfrak{a}$ we may set $z=0.$ What is left are only terms without a $z.$ None of them has $x_i^{m_i}$ in it if $m_i\geq d_i$ since those had been replaced by $z.$ We thus have only terms with lower exponents than $d_i.$ These count from $0,1,2,\ldots,d_i-1$ and are therefore $d_i$ many for each $i$.

$x_1^{m_1}\cdots x_n^{m_n}, m_i\leq d_i-1$ span $k[x_1,\ldots,x_n]/\mathfrak{a}$ which are at most $d_1\ldots d_n$ many of them, all other terms had a $z$ and vanished into the ideal $\mathfrak{a}.$

Also, how does the

"the classes $x_1^{m_1}\cdots x_n^{m_n}, m_i\leq d_i-1$"

span the vector space $k[x_1,\ldots,x_n]/\mathfrak{a}$?

Because every polynomial $P$ can be written this way after the $z=0$ procedure. You cannot find a polynomial in $k[x_1,\ldots,x_n]/\mathfrak{a}$ that is not composed of these monomials.

Finally, when the author chooses a particular monomial then says "there exists at least one $i$ such that $m_i\geq d_i$" make the proof to feel like he is going for a proof by contradiction. Since in the beginning of the proof, it assumes "for every $i$, now, there is at least one $i$.

He just describes that all $x_i^{m_i}$ with $m_i\geq d_i$ vanish into $\mathfrak{a}.$ That is the procedure going from $k[x_1,\ldots,x_n]$ to $k[x_1,\ldots,x_n]/\mathfrak{a}.$ The procedure I described with the auxiliary variable $z.$ It is what the ring homomorphism does, the transition from $x$ to $u.$ We have arbitrary powers of $x,$ but only powers $u^m$ with $m<d.$ The rest have gone into $\mathfrak{a}.$ It is therefore no proof by contradiction, it is what happens at
$$
\pi\, : \,k[x_1,\ldots,x_n]\longrightarrow k[x_1,\ldots,x_n]/\mathfrak{a}
$$
i.e.
\begin{align*}
\pi(P)&=\pi\left(\sum_{m_1,\ldots,m_n\geq 0}a_{m_1,\ldots,m_n} x_1^{m_1}\cdots x_n^{m_n}\right)\\[6pt]
&=\sum_{m_1,\ldots,m_n\geq 0}a_{m_1,\ldots,m_n} \pi(x_1)^{m_1}\cdots \pi(x_n)^{m_n}\\[6pt]
&=\sum_{m_1,\ldots,m_n\geq 0}a_{m_1,\ldots,m_n} u_1^{m_1}\cdots u_n^{m_n}\\[6pt]
&=\sum_{m_1<d_1,\ldots,m_n<d_n}a_{m_1,\ldots,m_n} u_1^{m_1}\cdots u_n^{m_n}
\end{align*}
because all $\pi(x_i^{m_i})=u_i^{m_i}=0$ if $m_i\geq d_i.$

The decomposition $x_i^{m_i}=x_i^{m_i-d_i}\cdot x_i^{d_i} \in \mathfrak{a}$ if $m_i-d_i\geq 0$ was only another way to say that those powers vanish into $\mathfrak{a}.$

These were three different ways to describe it: the author's method with $x_i,$ the trick with $z=0,$ and the formal method with the ring homomorphism $\pi(x_i)=u_i.$ But they are all the same. It's not an additional assumption in the proof, it is what happens if we say modulo $\mathfrak{a}.$

 

[elias001]

@fresh_42 I don't know why the person who came up with the answer in both direction did not put more details into the write up. I get the impression that he is writing for a journal article. I replied to the last post that you posted in my first thread I made on here a few days ago. I tried out the other direction and wrote it out. I was wondering if you can look it over and give some feedback. The post is here: https://www.physicsforums.com/threa...-x-is-irreducible-in-bbb-q-_-bbb-z-x.1080373/

 

[mathwonk]

I keep thinking this should be easy but don't actually see why. However, assuming the field is algebraically closed, it is a direct corollary of the most basic algebraic geometric result, namely the nullstellensatz. [I see it now. It is deduced in the last paragraph from an abstract algebraic version of the nullsatz, i.e. the fact that the radical of an ideal A equals the intersection of all prime ideals containing A.].
Geometric argument: One looks at the common zeroes V(A) of the monomials (equivalently all elements) in the ideal A, and notices that since all monomials are homogeneous, V(A) must form a cone, and indeed is a union of coordinate spaces of various dimensions. If I(V(A)) is the ideal of all polynomials vanishing on V(A), then the quotient k[X]/I(V(A)) is the ring of polynomial functions on V(A). Since A is contained in I(V(A)), and the quotient k[X]/A is already finite dimensional, so is the quotient k[X]/I(V(A)), so this set V(A) of common zeroes cannot contain any coordinate axis, hence equals only the origin.
By the nullstellensatz, the ideal I(V(A)) of all polynomials vanishing on the set V(A) (i.e. the intersection of all maximal ideals containing A) is just the radical of A, which solves the problem. I.e. since V(A) equals only the origin, the ideal I(V(A)) is the maximal ideal M = (X1,...,Xn), which thus equals the radical of A. Thus all elements Xj are in the radical of A, i.e. some power of each Xj belongs to A.

I keep trying some more algebraic argument using the fact that a finite dimensional domain is a field, hence all prime ideals containing A must be maximal. I don't quite see how to deduce that M is the only prime ideal containing A, which is what we need to deduce that rad(A) = M.....???

OK, how's this? By definition of a prime ideal, and the nature of a monomial, a minimal prime ideal P containing A is generated by some subset of the variables. But again, since k[X]/P is finite dimensional when P contains A, then P must actually contain all the variables, i.e. P=M = (X1,...,Xn). Then use fact that the radical of A ( = the ideal of those elements w such that some positive power of w lies in A), is equal to the intersection of all primes containing A; (this uses Zorn's lemma).

That 3 - sentence argument is valid for any field.

 

[mathwonk]

Now let me explain this solution in more detail, since it illustrates the power of abstract concepts. The problem asks us to show that if, for the ideal A:

1) A is generated by monomials, and

2) k[X]/A is finite dimensional over k,

then some positive power of every variable Xj belongs to A.

The first observation is that this property means we are asked to compute the “radical” of A. I.e. the radical of an ideal A consists of those elements x such that some positive power of x belongs to A. So we are asked to prove that under the two given hypotheses, that rad(A) equals the maximal ideal (X1,…,Xn). It is elementary to check that the radical of an ideal is also an ideal, and contains the original ideal.

GEOMETRIC ARGUMENT:
In algebraic geometry one thinks of polynomials as functions on n -space, and one studies an ideal A by looking at the set of points V(A) in n-space where all functions in the ideal A vanish simultaneously. I.e. p is in V(A) if and only if f(p) = 0 for all f in A. When the field k is algebraically closed, the first basic theorem, Hilbert’s nullstellensatz, or “zero set theorem”, says that two ideals have the same common vanishing locus if and only if they have the same radical. In particular the radical of A is the largest such ideal, hence equals the ideal I(V(A)) of all functions that vanish identically on V(A). I.e. when k is algebraically closed, then f is in I(V(A)) if and only if f(p) =0 for all p in V(A), if and only if f is in rad(A), i.e. iff some positive power of f belongs to A.

Since the maximal ideal (X1,…,Xn) is exactly the ideal of all functions vanishing on the origin, to show that rad(A) = (X1,…,Xn) it would suffice to show that the only common zero of A is the origin. Since A is generated by monomials, its common vanishing locus V(A) is an intersection of unions of hyperplanes, hence is a union of coordinate subspaces.
Since a function f is in I(V(A)) if and only if it equals zero on V(A), the quotient space k[X]/I(V(A)) is the space of all polynomial functions on V(A). Moreover since I(V(A)) contains A, the natural map k[X]/A —-> k[X]/I(V(A)) is surjective, so finite dimensionality of the first space forces finite dimensionality of the second.
But the space of polynomial functions on V(A) would contain all polynomials in Xj if V(A) contained the Xj axis, so the set V(A) is a union of coordinate subspaces that does not contain any axes, hence consists only of the origin. qed. at least in the case that k is algebraically closed.


ALGEBRAIC ARGUMENT:
However we want to give an argument that does not assume k algebraically closed. So we need another characterization of the radical of the ideal A. A standard result in commutative algebra is that rad(A) equals the intersection of all prime ideals containing A, where an ideal P is “prime” if and only if whenever a product fg belongs to P, then either f or g belongs to P. In particular maximal ideals are prime. I.e. P is prime if and only if k[X]/P is an integral domain, and M maximal implies k[X]/M is a field hence a domain. (Recall a ring is a domain iff fg=0 implies either f or g is 0.)

[Motivation: This “standard result” is an abstract version of the nullsatz. I.e. if we think of points as represented by the maximal ideals of functions vanishing on them, then a function f vanishes on the point corresponding to a maximal ideal M if and only if f belongs to M. (If k is algebraically closed then all maximal ideals do correspond to points.) So, if you sort this all out, the ideal of all functions vanishing on all the points of the set V(A) would be the intersection of maximal ideals containing A. If we expand our notion of “points” to include all maximal ideals, and also “fat points” which are dense in higher dimensional irreducible sets, and which correspond to prime ideals, then we would expect I(V(A)) to equal the intersection of all prime ideals containing A. So the abstract nullsatz says that rad(A) = intersection of all prime ideals containing A.]

Assuming the abstract nullsatz, it suffices to prove that under the two given hypotheses 1), 2), on A from the problem above, that (X1,…,Xn) is the only prime ideal containing A. Now by definition of prime ideal P, if P contains a monomial, i.e. a product of the variables, then P contains at least one of the variables. And if P contains even one of the variables in a monomial, then by definition of an ideal, P contains the monomial. Hence a prime ideal P contains an ideal A generated by monomials, if and only if P contains at least one variable from each generating monomial. Hence the minimal prime ideals P containing A (whose intersection is the same as that of all primes containing A) are ideals generated by some of the variables Xj, e.g. (X1,X4,X7). But now we use the fact that by property 2), we must also have k[X]/P finite dimensional whenever P contains A. Hence the only prime ideal P containing A must be the maximal ideal (X1,…,Xn). Thus rad(A) = (X1,…,Xn) as desired.

[For experts, to prove the abstract nullsatz, first prove that the ideal of nilpotent elements of a ring R consists of the intersection of all prime ideals of R, then note that the radical of A consists of the nilpotent elements of R/A.
It is elementary that all nilpotents of R belong to all primes of R. So if f is not nilpotent, it suffices to produce a prime P not containing f. Define the ring of fractions R_f, whose denominators are non negative powers of f, and consider the natural map R-->R_f, sending g to g/1. Since f is a unit in R_f, no proper ideals of R_f contain f. By Zorn's lemma R_f contains (proper) maximal, hence prime, ideals, and then we pull back one of them to R, getting a prime that does not contain f.]

 

[mathwonk]

Wait a minute,isn't this problem much easier than I have been making it look? I.e. suppose A is an ideal in k[X1,...,Xn] generated by monomials, i.e. products of some of the variables. Suppose no generating monomial contains X1. Then all generators are contained in the ideal (X2,...,Xn), hence k[X]/A ---> k[X]/(X2,...,Xn) ≈ k[X1] is surjective, so k[A] cannot be finite dimensional. Hence some monomial generator does contain X1. Now suppose every monomial generator containing X1 also contains some other variable. Then again every monomial generator is contained in the ideal (X2,...,Xn) and again A is contained in this ideal, contradicting finite dimensionality of k[X]/A. Hence some monomial generator contains X1 and only X1, i.e. some power of X1 is a generator, and A contains a power of X1. The same argument shows for all Xj, some power of Xj lies in A. I apologize for not reading this long thread, and presume this argument appears there somewhere, (unless I have made a naive mistake).

 

[fresh_42]

Wait a minute, isn't this problem entirely trivial?

Well, that depends on your level of education. It is nearly trivial for you and me, but not automatically trivial for students with little experience with quotient rings. And a proof can be very teaching, e.g., induction, the insertion homomorphism, and things like that.

By the way, that has puzzled me for years: do you say factor ring and factoring (out) or quotient ring and dividing (by)?

 

[mathwonk]

A one sentence summary: If A is an ideal of k[X] generated by monomials none of which is a power of Xj, then A is contained in the ideal (X1,...,Xj-1,Xj+1,...,Xn) and hence k[X]/A--->k[X]/(X1,...,Xj-1,Xj+1,...,Xn) ≈ k[Xj] is a surjection and k[X]/A could not be finite dimensional.

Thank you, let me rephrase; instead of "isn't this problem trivial", I should have said "isn't this problem much easier than I have been making it look?" My feeling though is that I think using induction also makes it look harder than it is. I agree however that for someone unfamiliar with quotient rings, induction might seem easier.

I call the rings k[X]/A either quotient rings or factor rings, usually the former. Zariski-Samuel and one other book of mine call them residue class rings, two of my books call them factor rings, one calls them difference rings, and 8 of my other books call them quotient rings.

 

[fresh_42]

I am uncertain whether it helps to understand the concept or if it adds confusion, but the whole situation can be translated to numbers. We could choose $k=\mathbb{Q}$ and instead of indeterminates, we choose transcendental numbers, e.g. $\mathbb{Q}\left[\pi, e, 2^\sqrt{2}\right].$ That makes it look a bit more familiar. The downside is that it is not obvious that the three numbers are $\mathbb{Q}$-linearly independent.

 

[elias001]

@mathwonk I just saw your replies. The book where this question came from only assumes the reader has basic linear algebra and a semester of abstract algebra. In the latter, the reader is not assumed to know anything about basics of Galois theory. The high brow machinery you are using get introduced later in the text. I am about to email the author and ask some questions relating to the exercise. I will get back to you and @fresh_42 what he has to say.

Also i saw somewhere that you worked with Thomas Finney before. His calculus text was the one I used for self teaching myself Calculus.